If dim\(M=0\) , then \(M\cong Q/(y^{i(M)})\oplus \ldots \oplus Q/(y^{i(M)})\oplus Q/(y^{i(M)+1}) \) , because \(e(M)=\mu (M)i(M)+1\)  (see [1]}). This implies \(h_{M}(z)=\mu (M)(1+z+\ldots +z^{i(M)-1})+z^{i(M)}\) .
If dim\(M=1\) , then from REF (1) we have
\(h_{M}(z)=\mu (M)(1+z+\ldots +z^{i(M)-1})+z^{s}\)  for \(s\ge i(M)\) , because \(e(M)=i(M)\mu (M)+1\) . Let \(x_1\)  be a \(\phi \) -superficial element. Set \(M_1=M/x_1M\) . We know that \(e(M)=e(M_1)\) , \(\mu (M)=\mu (M_1)\)  and \(i(M)=i(M_1)\) . So from the dimension zero case \(h_{M_1}(z)=\mu (M)(1+z+\ldots +z^{i(M)-1})+z^{i(M)}\) . Therefore \(G(M)\)  is Cohen-Macaulay if and only if \(h_{M_1}(z)=h_M(z)\)  (see REF ) if and only if \(s=i(M)\) .
If dim\(M=2\) .
Let \(\underline{x}=x_1,x_2\)  be a maximal \(\phi \) -superficial sequence (see REF ). Set \(M_1=M/x_1M\) , \(M_2=M/\underline{x}M\) , \(J=(x_1,x_2)\)  and \((Q^{\prime },(y))=(Q/(\underline{x}),\mathfrak {n}/(\underline{x}))\) .
Clearly, \(Q^{\prime }\)  is a DVR. Notice that since \(\underline{x}\)  is \(\phi \) -superficial sequence, \(e(M)=e(M_1)=e(M_2),\mu (M)=\mu (M_1)=\mu (M_2)\)  and \(i(M)=i(M_1)=i(M_2)\) . So \(M_2\cong Q^{\prime }/(y^{i(M)})\oplus \ldots \oplus Q^{\prime }/(y^{i(M)})\oplus Q^{\prime }/(y^{i(M)+1}) \) , because \(e(M)=\mu (M)i(M)+1\) . This implies \(h_{M_2}(z)=\mu (M)(1+z+\ldots +z^{i(M)-1})+z^{i(M)}\) .
Since dim\(M_1=1\) , \(e(M)=e(M_1)\) , \(\mu (M)=\mu (M_1)\)  and \(i(M)=i(M_1)\) , we get \(e(M_1)=\mu (M_1)i(M_1)+1\) . So from dimension one case, \(h-\) polynomial of \(M_1\)  is
\(h_{M_1}(z)=\mu (M)(1+z+\ldots +z^{i(M)-1})+z^{s}\)  for \(s\ge i(M)\) .
Since \(M=coker(\phi )\) , for \(n\le i(M)-1\)  we get
\(\mathfrak {m}^nM/{\mathfrak {m}^{n+1}M}\cong \mathfrak {m}^n(Q)^{\mu (M)}/{\mathfrak {m}^{n+1}(Q)^{\mu (M)}}. \) 
Now if \(h\) -polynomial of \(M\)  is \(h_M(z)=h_0(M)+h_1(M)z+\ldots +h_t(M)z^t\)  then
\(\ell (\mathfrak {m}^nM/\mathfrak {m}^{n+1}M)=\binom{n+2}{n}\mu (M)\ \text{for all}\ n\le i(M)-1.\) 
So for all \( n\le i(M)-1\)  we have
\((n+1)h_0(M)+nh_1(M)+\ldots +h_n(M)=\binom{n+2}{n}\mu (M). \qquad \mathrm {(\dagger )}\) 
Now since \(h_0(M)=\mu (M)\) , we get from (\(\dagger \) )
\(h_0(M)=h_1(M)=\ldots =h_{i(M)-1}(M)=\mu (M)\) 
So we have, \(h_n(M)=h_n({M_1})\)  for all \(n\le i(M)-1\) .
From Singh's equality (REF ) we have
\(\mathfrak {m}^{n+1}M:x_1=\mathfrak {m}^nM\ \text{for}\ n=0,\ldots ,i(M)-1.\) 
So we have
\(\mathfrak {m}^{n+1}M\cap x_1M=x_1\mathfrak {m}^nM\ \text{for}\ n=0,\ldots ,i(M)-1\) 
Since \(h_n(M_1)= h_n(M_2)\)  for \(n=0,\ldots ,i(M)-1\) ,
from Singh's equality (REF )
\(\mathfrak {m}^{n+1}{M_1}:x_2=\mathfrak {m}^n{M_1} \ \text{for}\ n=0,\ldots ,i(M)-1\) 
Now we have \(\mathfrak {m}^{n+1}M\cap JM=J\mathfrak {m}^nM\ \text{for}\ n=0,\ldots ,i(M)-1\) .
In fact, if \(\alpha =ax_1+bx_2\in \mathfrak {m}^{n+1}M\) . Going modulo \(x_1\)  we get \(\overline{\alpha }=\overline{b}x_2\in \mathfrak {m}^{n+1}{M_1}\) . From (REF ) we have \(\overline{b}\in \mathfrak {m}^n{M_1}\) . So we can write \(b=c+f\) , where \(c\in \mathfrak {m}^nM\)  and \(f\in x_1M\) . This implies \(\alpha =ax_1+cx_2+fx_2\) . Hence \(\alpha -cx_2\in \mathfrak {m}^{n+1}M\cap x_1M\) . So from (REF ) we have \(\alpha =cx_2+gx_1\)  with \(c,g\in \mathfrak {m}^nM\) . This implies \(\alpha \in J\mathfrak {m}^nM\) .
So we have
\(vv_i(M)=\ell \Big (\frac{\mathfrak {m}^{i+1}M\cap JM}{J\mathfrak {m}^iM}\Big )=0 \ \text{for}\ i=0,\ldots ,i(M)-1\) 
Since \(\mathfrak {m}^{i(M)}M:x_1=\mathfrak {m}^{i(M)-1}M\) , we have from  REF
\(v_{i(M)-1}=\ell (\mathfrak {m}^{i(M)}M/J\mathfrak {m}^{i(M)-1}M)=\ell (\mathfrak {m}^{i(M)}{M_1}/x_2\mathfrak {m}^{i(M)-1}{M_1})=\rho _{i(M_1)-1}= 1\) 
Notice that last equality in REF  is clear from the \(h\) -polynomial of \(M_1\) . In fact, from the expression of \(h_{M_1}\)  and REF  we get \(\rho _{i(M_1)-1}(M_1)=\rho _{i(M_1)}(M_1)=\ldots =\rho _{s-1}(M_1)=1\) . Here \(\rho _n(M_1)=\ell (\mathfrak {m}^{n+1}M_1/x_2\mathfrak {m}^nM_1)\)  for all \(n\ge 0\) .
Now from conditions (REF ) and (REF ), depth\(G(M)\ge 1\)  (see [2]}, take \(p=i(M)-1\) ). So \(h_M(z)=h_{M_1}(z)\)  (see REF ). Also, \(G(M)\)  is Cohen-Macaulay if and only if \(s=i(M)\) .
Now assume dim\(M\ge 3\)  and \(\underline{x}=x_1,\ldots ,x_d\)  a maximal \(\phi \) -superficial sequence. Set \(M_{d-2}=M/(x_1,\ldots ,x_{d-2})M\) . So, depth\(G(M_{d-2})\ge 1\) .
By Sally-descent we get depth\(G(M)\ge d-1\)  and \(h_M(z)=\mu (M)(1+z+\ldots +z^{i(M)-1})+z^s\)  where \(s\ge i(M)\) . Note that \(G(M)\)  is Cohen-Macaulay if and only if \(s=i(M)\) .
